Problem: $\begin{aligned} y&=\sin(8-9x) \\\\ \dfrac{dy}{dx}&=\,? \end{aligned}$ Choose 1 answer: Choose 1 answer: (Choice A) A $\sin(-9)$ (Choice B) B $-9\cos(8-9x)$ (Choice C) C $-9\cos(x)$ (Choice D, Checked) D $(9x-8)\sin(x)$
Answer: Since $\sin(8-9x)$ is a composite function, we can use the chain rule. The chain rule says: $\dfrac{d}{dx}\left[w\Bigl(u(x)\Bigr)\right]={w'\Bigl({u(x)}\Bigr)}\cdot{u'(x)}$ To use it, we need to recognize our function as a composite function like $w\big(u(x)\big)$. $\underbrace{\sin(~\overbrace{8-9x}^{\text{inner}}~)}_{\text{outer}}$ So if $\sin(8-9x)=w(u(x))$, then: $\begin{aligned} {u(x)}&={8-9x} &&\text{inner function} \\\\ w(x)&=\sin(x)&&\text{outer function} \end{aligned}$ Before applying the chain rule, let's find the derivatives of the inner and outer functions (work not shown, but hopefully these derivatives are familiar by now). $\begin{aligned} {u'(x)}&={-9} \\\\ {w'(x)}&={\cos(x)} \end{aligned}$ Now let's apply the chain rule: $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{d}{dx}\left[w\Bigl(u(x)\Bigr)\right] \\\\ &={w'\Bigl({u(x)}\Bigr)}\cdot{u'(x)} \\\\ &={\cos({8-9x})} \cdot {-9} \\\\ &=-9\cos(8-9x) \end{aligned}$